Given the ellipse $4x^2+9y^2=36 \rightarrow \large\frac{x^2}{9}$$+\large\frac{y^2}{4}$$=1$

Since $9 \gt 4$, he major axis is along the y-axis.

On comparing the equation with $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1 \rightarrow a = \sqrt {9} = 3$ and $b = \sqrt {4} =2$

$\Rightarrow c = \sqrt {a^2-b^2} = \sqrt{9-4} = \sqrt{5}$

1) The coordinates of the foci are $(\sqrt{ 5}, 0)$ and $(-\sqrt{ 5}, 0)$

2) The coordinates of the vertex are $(3,0)$ and $(-3,0)$

3) The Length of the major axis, $2a = 2 \times 3= 6$

4) The Length of the minor axis, $2b = 2 \times 2= 4$

5) The Eccentricity, $e = \large\frac{c}{a} $$ = \large\frac{\sqrt{5}}{3}$

6) Length of Latus Rectum, $\large\frac{2b^2}{a} $$ = \large\frac{2 \times 2^2}{3} $$ = \large\frac{8}{3}$